H mole of Iron liberates mole of Hydrogen gas . g Iron = mole Iron ∴ n = T = C = K w = – P D V w P nRT P = − w = – nRT w = – × . × J w = – . J w = – .
kJ . T i = C = K T f = C = K 11th Std Chemistry 11th Std Chemistry - - - - D H = nC p (T f – T i ) ∆ H R ( ) D H = – R . C + O → CO D H = – a KJ ....................... (i) 2CO + O 2CO D H = –b kJ .......................
(ii) C O CO H → ∆ ? (i) × 2C + 2O → 2CO D H = – 2a kJ ....................... (iii) Reverse of equation (ii) will be 2CO → 2CO + O D H = + b kJ ....................... (iv) (iii) + (iv) 2C + O → 2CO D H = b – 2a kJ .......................
(v) (v) ÷ C O CO H b a kJ → ∆ ( ) . Given : D H C (CH ) = – kJ mol – D H C (C H ) = – kJ mol – Let the mixture contain lit of and lit of propane. CH + 2O → CO + 2H O x 2x C H + 5O → 3CO + 4H O ( . – x) ( .
– x) Volume of oxygen consumed = 2x + ( . – x) = lit 11th Std Chemistry 11th Std Chemistry - - - - 2x + . – 5x